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Hint: Here we solve the problem by finding the given set is a function or not.

Given $g = \{ (1,1),(2,3),(3,5),(4,7)\} $

Here, if we observe in the given set each element of domain has unique image so from this we can say the g is function.

Here the function g is described by a formula $g(x) = \alpha x + \beta $

Now let us substitute $x = 1$ in the formula $g(x) = \alpha x + \beta $

$\

g(1) = \alpha (1) + \beta \\

g(1) = \alpha + \beta \to 1 \\

\ $

We know that $g(1) = 1$ from the given function g

So we write the equation as

$\alpha + \beta = 1 \to 1$

Now again let us substitute $x = 2$in the formula $g(x) = \alpha x + \beta $

$\

g(2) = \alpha (2) + \beta \\

g(2) = 2\alpha + \beta \\

\ $

And we know that $g(2) = 3$from the given function g

So we can write the equation as

$2\alpha + \beta = 3$$ \to 2$

Now let us solve equation 1 and 2

We get $\alpha = 2$ and $\beta = 1$

Now we can write the formula as $g(x) = 2x - 1$

$\therefore g(x) = 2x - 1$

NOTE: We have used the domain values of g function to get the formula of g(x).

Given $g = \{ (1,1),(2,3),(3,5),(4,7)\} $

Here, if we observe in the given set each element of domain has unique image so from this we can say the g is function.

Here the function g is described by a formula $g(x) = \alpha x + \beta $

Now let us substitute $x = 1$ in the formula $g(x) = \alpha x + \beta $

$\

g(1) = \alpha (1) + \beta \\

g(1) = \alpha + \beta \to 1 \\

\ $

We know that $g(1) = 1$ from the given function g

So we write the equation as

$\alpha + \beta = 1 \to 1$

Now again let us substitute $x = 2$in the formula $g(x) = \alpha x + \beta $

$\

g(2) = \alpha (2) + \beta \\

g(2) = 2\alpha + \beta \\

\ $

And we know that $g(2) = 3$from the given function g

So we can write the equation as

$2\alpha + \beta = 3$$ \to 2$

Now let us solve equation 1 and 2

We get $\alpha = 2$ and $\beta = 1$

Now we can write the formula as $g(x) = 2x - 1$

$\therefore g(x) = 2x - 1$

NOTE: We have used the domain values of g function to get the formula of g(x).

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